Combinations?

This is the problem straight off my homework:


"In a bag are 15 red marbles and 10 green marbles. 5 marbles are randomly chosen. How many combinations of marbles can be drawn from the bag at random if at least 4 red marbles are drawn?"


My original answer for the solution was





15! / (11! * 4!) * 21! / (20! * 1!) = 28665


(Which is also C(15, 4) * C(21, 1))





I ended up being wrong, with the answer actually being





15! / (10! * 5!) + 15! / (11! * 4!) * 10! / (9! * 1!) = something that isn't 28665


(Which is C(15,5) + C(15,4) * C(10,1))





I understand how the second answer is correct. It's just the combinations of 4 red marbles + 1 green marble added to the combinations of 5 red marbles. What I don't understand is why my original answer is wrong. As far as I know, I took the number of combinations involving 4 red marbles and any one of the remaining 21 marbles. Could somebody offer an explanation of this to me?

Combinations?
Your answer splits the sample of 5 marbles into two pieces: the first 4 marbles, which must be red, and the last marble, which can be red or green. If the last marble is green this is unproblematic, but if the last marble is red this means that each combination of marbles gets counted five times, as if you draw red marbles 1, 2, 3, 4, and 5 from the bag, either marble 1, marble 2, ..., or marble 5 could be counted as the last marble. So, to fix your factor of C(21,1) = 21 = 10 + 11, the 10 may be left as is but the 11 should be divided by 5, giving an answer of


C(15, 4) * (10 + 11/5) = 16653


which is correct as it is equal to


C(15,5) + C(15,4)*C(10,1) = 16653.
Reply:because in your original answer, we double (more of quintuple actually) counted.





(4 of 15 reds) + (1 of 21 remaining)


=%26gt; (r1, r2, r3, r4) + (drawn a r5)





but there's also other 4 equivalent combination that you had also counted in ;


=%26gt; (r1, r2, r3, r5) + (drawn a r4)


=%26gt; (r1, r2, r4, r5) + (drawn a r3)


=%26gt; (r1, r3, r4, r5) + (drawn a r2)


=%26gt; (r2, r3, r4, r5) + (drawn a r1)





i myself always did it like your original answer. i really need to remind myself this.